Let $F$ be the center of the division ring $D$. Then $D$ represents its class in the Brauer group $Br(F)$. The opposite ring $D^{opp}$ represents the inverse element. The reason why for example the quaternions are isomorphic to their opposite algebra is that the quaternions are an element of order 2 in $Br(\mathbf{R})$, and hence equal to its own inverse in the Brauer group.
To get a division algebra that is not isomorphic to its opposite algebra we can use an element of order 3 in the Brauer group. One method for constructing those is to start with a (cyclic) Galois extension of number fields $E/F$ such that $[E:F]=3$. Let $\sigma\in Gal(E/F)$ be the generator. Let $\gamma\in F$ be an element that cannot be written in the form $\gamma=N(x)$, where $N:E\rightarrow F, x\mapsto x\sigma(x)\sigma^2(x)$ is the relative norm map. Consider the set of matrices
$$
\mathcal{A}(E,F,\sigma,\gamma)=\left\{
\left(\begin{array}{rrr}
x_0&\sigma(x_2)&\sigma^2(x_1)\\
\gamma x_1&\sigma(x_0)&\sigma^2(x_2)\\
\gamma x_2&\gamma\sigma(x_1)&\sigma^2(x_0)
\end{array}\right)\mid x_0,x_1,x_2\in E\right\}.
$$
A theorem of A. Albert tells us that this forms a division algebra with center $F$, and its order in $Br(F)$ is 3, so it will not be isomorphic to its opposite algebra. The theory is described for example in ch. 8 of Jacobson's Basic Algebra II. The buzzword 'cyclic division algebra' should give you some hits.
For a concrete example consider the following. Let $F=\mathbf{Q}(\sqrt{-3})$ and let
$E=F(\zeta_9)$, with $\zeta_9=e^{2\pi i/9}$. Then $E/F$ is a cubic extension of cyclotomic fields, $\sigma:\zeta_9\mapsto\zeta_9^4$. I claim that the element $2$ does not belong the image of the norm map. This follows from the fact 2 is totally inert in the extension tower $E/F/\mathbf{Q}$. Basically because $GF(2^6)$ is the smallest finite field of characteristic 2 that contains a primitive ninth root of unity. Now, if $2=N(x)$ for some $x\in E$, then 2 must appear as a factor (with a positive coefficient) in the fractional ideal generated by $x$. But the norm map then multiplies that coefficient by 3, and as there were no other primes above 2, we cannot cancel that. Sorry, if this is too sketchy.
Anyway (see Jacobson again), the product of the $\gamma$ elements modulo $N(E^*)$ is the operation in the Brauer group $Br(E/F)\le Br(F).$ Therefore the opposite algebra should correspond to the choice $\gamma=1/2$, (or to the choice $\gamma=4$, as $2\cdot4=N(2)$). So
$$
\mathcal{A}(E,F,\sigma,2)^{opp}\cong
\mathcal{A}(E,F,\sigma,1/2).
$$
and the choices $\gamma=2$ and $\gamma=1/2$ yield non-isomorphic division algebras, as their ratio $=4$ is not in the image of the norm map.
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Edit (added more details here, because another answer links to this answer): In general the cyclic division algebra construction works much the same for any cyclic extension $E/F$. When $[E:F]=n$ we get a set of $n\times n$ matrices with entries in $E$. The number $\gamma$ appears in the lower diagonal part of the matrix. The condition for this to be a division algebra is that $\gamma^k$ should not be a norm for any integer $k, 0<k<n$. Obviously it suffices to check this for (maximal) proper divisors of $n$. In particular, if $n$ is a prime, then it suffices to check that $\gamma$ itself is not a norm.
Edit^2: Matt E's answer here linear algebra over a division ring vs. over a field gives a simpler cyclic division algebra of $3\times3$ matrices with entries in the real subfield of the seventh cyclotomic field.
No, not only is there not any "natural" or "canonical" way to make it into a functor, but there is actually no functor at all which on objects sends each abelian group to its ring of endomorphisms. For instance, there exists a homomorphism between any two abelian groups (the zero homomorphism), so to have such a functor there would need to exist a homomorphism between any two endomorphism rings of abelian groups. This is not true (for instance, consider $\mathbb{Z}/(p)$ and $\mathbb{Z}/(q)$ for distinct primes $p$ and $q$).
Best Answer
The question for which rings $R$ there is some isomorphism $R \cong \mathrm{End}_{\mathbb{Z}}(U(R))$ (where $U(R)$ denotes the underlying $\mathbb{Z}$-module) is rather unnatural and I bet that we cannot answer it.
It is more natural to ask for which rings $R$ the canonical homomorphism $\lambda : R \to \mathrm{End}_{\mathbb{Z}}(U(R))$ which maps $r \in R$ to $(x \mapsto rx)$ is an isomorphism. I will try to answer this.
Note that there is a homomorphism $\rho : U(\mathrm{End}_{\mathbb{Z}}(U(R))) \to U(R)$ mapping $f \mapsto f(1)$ (if these $U$'s confuse you, just forget about them; I am just saying that $\rho$ is additive, not multiplicative in general), and we have $\rho \circ U(\lambda) = \mathrm{id}_{U(R)}$. Thus, $\lambda$ is a monomorphism, and it is an isomorphism if and only if $U(\lambda) \circ \rho = \mathrm{id}$. And this means, by definition: If $\alpha : U(R) \to U(R)$ is a homomorphism, then $\alpha(x)=\alpha(1) x$ for all $x \in R$. In other words, $\alpha$ is $R$-linear (where $R$ acts on the right).
This is easily seen to be equivalent to the following: If $M$ is some right $R$-module, then any $\mathbb{Z}$-linear map $M \to R$ (I ignore the $U$s here) is automatically $R$-linear. Now let us look at a stronger property: If $M,N$ are any two right $R$-modules, then any $\mathbb{Z}$-linear map $M \to N$ is already $R$-linear. It turns out (by formal nonsense) that this property is equivalent to the condition that $\mathbb{Z} \to R$ is an epimorphism in the category of rings. One can classify these rings in the commutative case - Torsten Schöneberg has given a nice overview here:
If $R$ is a commutative ring, and $\mathbb{Z} \to R$ is an epi which is not injective, then $R=\mathbb{Z}/n$ for some $n \in \mathbb{Z}$. If it is injective, then there is a set of primes $\widehat{P}$, a subset $P$ and a function $n : P \to \mathbb{N}^+$, such that $R$ is isomorphic to the (possibly infinite) tensor product of the $(\widehat{P} \setminus P)^{-1}$-algebras $\mathbb{Z}/p^{n(p)} \times p^{-1} \mathbb{Z}$, where $p \in P$. Thus, the epimorphisms are built up canonically out of a) quotients, b) localizations, c) tensor products, d) directs products $R \times S$ with $\mathbb{Z} \to R,\,S$ epi such that $R \otimes S=0$.